Question: You have found the following ages (in years) of all 4 zebras at your local zoo: $ 14,\enspace 15,\enspace 9,\enspace 1$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{14 + 15 + 9 + 1}{{4}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $14$ years $4.2$ years $17.64$ years $^2$ $15$ years $5.2$ years $27.04$ years $^2$ $9$ years $-0.8$ years $0.64$ years $^2$ $1$ year $-8.8$ years $77.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{17.64} + {27.04} + {0.64} + {77.44}} {{4}} $ $ {\sigma^2} = \dfrac{{122.76}}{{4}} = {30.69\text{ years}^2} $ The average zebra at the zoo is 9.8 years old. The population variance is 30.69 years $^2$.